The probability of selecting 6 and leaving 2 such that 1 is white and the other blue is the same as the probability of selecting 2 such that one is white and the other blue while 6 are remaining.

Probability of picking first blue and then white = (3/8)*(5/7)
Probability of picking first white and then blue = (5/8)*(3/7)
Total probability = (3/8)*(5/7) + (5/8)*(3/7) = 15/28

Answer (D)
_________________

Good question. +1

The required probability = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls.

Ways to select 6 out of total 8 = 8C6

Ways to select 4 out of 5 white balls = 5C4

Ways to select 2 out of 3 blue balls = 3C2

Thus the required probability = (5C4*3C2)/8C6 = 15/28.

D is thus the correct answer.
_________________

Hi,
another way to look at the problem..

what is the way to have 1 of Blue and 1 of white in the last two remaining..
it is to pick only 2 BLUE in the first 6 picked
WAYs to have 2 BLUE in the 6 picked = 6C2= 15..
what are total ways to pick 6 out of 8 = 8C6= 28..
The prob = 15/28
D

Hello VeritasPrepKarishma, can you explain how this is the same.

It is true because the total number of balls of both the colors does not change. It does not matter whether you reach 1 white ball at the end by removing 4 white before or you remove 1 white to have 4 white remaining.

In simpler terms , Ill use the example fo the white balls: you end up getting the same answer whether you do 1+4 or 4+1.

If you remove 1 w and 1 b ball at the start, the probability = (5C1*3C1)/8C6

If you remove 1 w and 1 b ball at the end, the probability = (5C4*3C2)/8C6

Both these equations give the same value as 5C1 = 5C4 and 3C1 = 3C2

Hope this helps.
_________________

Honestly the easiest way to solve this is to draw it out and eyeball it.

B B B W W W W W

Visually, you can see that there are 15 combinations of one blue marble and one white marble (the first "B" combined with all 5 "W" + the second "B" combined with all 5 "W" + the third "B" combined with all 5 "W" = 15).

Then just use the combination formula for 8C2 to find the total number of combinations.

8!/2!*6! = 8*7/2 = 28

Answer is thus 15/28

ok thanks. Looks like my quant concepts are quite rusty. Long way ahead...looks like .

Let us try to make this simpler. If the question had instead asked:

"A bag contains 3 blue and 5 white marbles. One by one, marbles are drawn out randomly until only ONE is left in the bag. What is the probability that it is blue?"

We can say that there are two equivalent ways of going about this:
1. We take out 7 marbles and leave one in the bag.
2. We take out one and leave 7 in the bag.

These two operations are equivalent. After all, we are randomly separating one marble from the other seven. Hence, these two experiments will yield the same probability.

You can now extend this to the case where 2 marbles need to be separated from the rest.

Ways to select out of total 8 = 8C6

Ways to select 4 out of 5 white balls = 5C4

Ways to select 2 out of 3 blue balls = 3C2

Thus the required probability = (5C4*3C2)/8C6 = 15/28.

D SHOULD BE THE ANSWER , PLEASE GIVE KUDOS, IF U LIKE .THANKS

oh damn...can't believe I missed this one...
overcomplicated myself...
8C6=28
5C4 * 3C2 = 5x3 = 15
15/28

damn...

Really like this approach. Just one question - do we consider the number of ways of choosing the balls? How can we decide when to do that? Please suggest.!
Thank you!!

Another quick solution:

Prob. that 1B & 1W remains in the bag = (Number of ways we can select 6 balls -2B & 4W - from the Bag) / (Total number of ways we can choose 6 balls from the bag)

Numerator = 6P6 / (2!*4!) = 15
Denominator = 8C6 = 28
Required prob. = 15/28

Thanks.

What do you mean by "do we consider the number of ways of choosing the balls", are you referring to the very process of choosing lets say 6 out of 8 balls or something else? Or is your question more about when to use the order of selection of balls (permutation vs combination)?
_________________

We need to determine the probability of selecting 2 blue marbles and 4 white marbles in 6 selections. The number of ways to select the blue marbles is 3C2 = 3 and the number of ways to select the white marbles is 5C4 = 5. Thus, the total number of ways is 3 x 5 = 15.

The total number of ways to select 6 marbles from 8 is 8C6 = (8 x 7 x 6 x 5 x 4 x 3)/6! = (8 x 7 x 6 x 5 x 4 x 3)/(6 x 5 x 4 x 3 x 2) = (8 x 7)/2 = 28.

Thus, the probability is 15/28.

Answer: D
_________________

There are two marbles remaining (one white and one blue). In other words, 2 blue marbles out of 3 and 4 white marbles out of 5 are remaining.
Ways to select 2 blue marbles: 3c2
Ways to select 4 white marbles: 5c4
Total ways to select is to select 6 marbles out of 8 i.e. 8c6
Probability: 3c2*5c4/8c6= 15/28

Another way to think of it...

If you arrange the 3 blue and 5 white in a row, how many of those arrangements end in BLUE-WHITE or WHITE-BLUE?

Total # of arrangements of 3 blue, 5 white: 8!/3!5! = 56
# of arrangements ending in B-W or W-B = (arrange 2 blue and 4 white) * 2 = (6!/2!4!) * 2 = 30

So, 30/56 (15/28) ways to arrange 3 blue and 5 white in a row such that the last 2 marbles are B-W or W-B.

Answer: D

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________